\section{向量组的线性表示}

	\begin{titwo}
		设向量组
		\[
			(\text{\Rmnum{1}})\bm \alpha_{1}, \bm \alpha_{2}, \cdots, \bm \alpha_{s}
		\]
		线性无关，(\Rmnum{2})$\bm \beta_{1}, \bm \beta_{2}, \cdots, \bm \beta_{t}$ 线性无关，且 $\bm \alpha_{i}(i = 1,2,\cdots,s)$ 不能由(\Rmnum{2})$\bm \beta_{1}, \bm \beta_{2}, \cdots, \bm \beta_{t}$ 线性表出，$\bm \beta_{j}(j = 1,2,\cdots,t)$ 不能由(\Rmnum{1})$\bm \alpha_{1}, \bm \alpha_{2}, \cdots, \bm \alpha_{s}$ 线性表出，则向量组 $\bm \alpha_{1}, \bm \alpha_{2}, \cdots, \bm \alpha_{s}, \bm \beta_{1}, \bm \beta_{2}, \cdots, \bm \beta_{t}$\kuo.

		\onech{必线性相关}{必线性无关}{可能线性相关，也可能线性无关}{以上都不正确}
	\end{titwo}

	\begin{titwo}
		设
		\begin{gather*}
			\bm \alpha_{1} = [1,0,-1,2]^{\TT}, \bm \alpha_{2} = [2,-1,-2,6]^{\TT},\\
			\bm \alpha_{3} = [3,1,t,4]^{\TT}, \bm \beta = [4,-1,-5,10]^{\TT},
		\end{gather*}
		已知 $\bm \beta$ 不能由 $\bm \alpha_{1}, \bm \alpha_{2}, \bm \alpha_{3}$ 线性表出，则 $t = $\htwo.
	\end{titwo}

	\begin{titwo}
		已知
		\begin{gather*}
			\bm \alpha_{1} = [1,-1,1]^{\TT},\bm \alpha_{2} = [1,t,-1]^{\TT},\\
			\bm \alpha_{3} = [t,1,2]^{\TT},\bm \beta = \bigl[4,t^{2},-4\bigr]^{\TT},
		\end{gather*}
		若 $\bm \beta$ 可由 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{3}$ 线性表示，且表示法不唯一，求 $t$ 及 $\bm \beta$ 的表达式.
	\end{titwo}

	\begin{titwo}
		已知 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{3},\bm \alpha_{4}$ 为 $3$ 维非零列向量，则下列结论：\\
		\circled{1}如果 $\bm \alpha_{4}$ 不能由 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{3}$ 线性表出，则 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{3}$ 线性相关;\\
		\circled{2}如果 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{3}$ 线性相关，$\bm \alpha_{2},\bm \alpha_{3},\bm \alpha_{4}$ 线性相关，则 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{4}$ 也线性相关;\\
		\circled{3}如果 $r(\bm \alpha_{1}, \bm \alpha_{1} + \bm \alpha_{2}, \bm \alpha_{2} + \bm \alpha_{3}) = r(\bm \alpha_{4}, \bm \alpha_{1} + \bm \alpha_{4}, \bm \alpha_{2} + \bm \alpha_{4}, \bm \alpha_{3} + \bm \alpha_{4})$，则 $\bm \alpha_{4}$ 可以由 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{3}$ 线性表出.\\
		其中正确的个数为\kuo.

		\fourch{$0$}{$1$}{$2$}{$3$}
	\end{titwo}

	\begin{titwo}
		向量组(\Rmnum{1}) $\bm \alpha_{1}, \bm \alpha_{2}, \cdots, \bm \alpha_{s}$，其秩为 $r_{1}$，向量组(\Rmnum{2}) $\bm \beta_{1}, \bm \beta_{2}, \cdots, \bm \beta_{s}$，其秩为 $r_{2}$，且 $\bm \beta_{i} (i = 1,2,\cdots,s)$ 均可由向量组(\Rmnum{1})$\bm \alpha_{1}, \bm \alpha_{2}, \cdots, \bm \alpha_{s}$ 线性表出，则必有\kuo.

		\onech{$\bm \alpha_{1} + \bm \beta_{1}, \bm \alpha_{2} + \bm \beta_{2}, \cdots, \bm \alpha_{s} + \bm \beta_{s}$ 的秩为 $r_{1} + r_{2}$}{$\bm \alpha_{1} - \bm \beta_{1}, \bm \alpha_{2} - \bm \beta_{2}, \cdots, \bm \alpha_{s} - \bm \beta_{s}$ 的秩为 $r_{1} - r_{2}$}{$\bm \alpha_{1}, \bm \alpha_{2}, \cdots, \bm \alpha_{s}, \bm \beta_{1}, \bm \beta_{2}, \cdots, \bm \beta_{s}$ 的秩为 $r_{1} + r_{2}$}{$\bm \alpha_{1}, \bm \alpha_{2}, \cdots, \bm \alpha_{s}, \bm \beta_{1}, \bm \beta_{2}, \cdots, \bm \beta_{s}$ 的秩为 $r_{1}$}
	\end{titwo}

	\begin{titwo}
		已知向量组
		\[
			\bm \alpha_{1} = \begin{bsmallmatrix}
				1 \\
				-1 \\
				2
			\end{bsmallmatrix},
			\bm \alpha_{2} = \begin{bsmallmatrix}
				0 \\
				3 \\
				1
			\end{bsmallmatrix},
			\bm \alpha_{3} = \begin{bsmallmatrix}
				3 \\
				0 \\
				7
			\end{bsmallmatrix}
		\]
		与向量组
		\[
			\bm \beta_{1} = \begin{bsmallmatrix}
				1 \\
				-2 \\
				2
			\end{bsmallmatrix},
			\bm \beta_{2} = \begin{bsmallmatrix}
				2 \\
				1 \\
				5
			\end{bsmallmatrix},
			\bm \beta_{3} = \begin{bsmallmatrix}
				x \\
				3 \\
				3
			\end{bsmallmatrix}
		\]
		等秩，则 $x = $\htwo.
	\end{titwo}

	\begin{titwo}
		已知 $\bm \alpha_{1} = [1,2,-3,1]^{\TT}, \bm \alpha_{2} = [5,-5,a,11]^{\TT}, \bm \alpha_{3} = [1,-3,6,3]^{\TT}, \bm \alpha_{4} = [2,-1,3,a]^{\TT}$. 问：
		\begin{enumerate}
			\item 当 $a$ 为何值时，向量组 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{3},\bm \alpha_{4}$ 线性相关;
			\item 当 $a$ 为何值时，向量组 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{3},\bm \alpha_{4}$ 线性无关;
			\item 当 $a$ 为何值时，$\bm \alpha_{4}$ 能由 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{3}$ 线性表出，并写出它的表出式.
		\end{enumerate}
	\end{titwo}

	\begin{titwo}
		已知
		\[
			\bm \alpha_{1} = \begin{bsmallmatrix}
				1 + \lambda \\
				1 \\
				1
			\end{bsmallmatrix},
			\bm \alpha_{2} = \begin{bsmallmatrix}
				1 \\
				1 + \lambda \\
				1
			\end{bsmallmatrix},
			\bm \alpha_{3} = \begin{bsmallmatrix}
				1 \\
				1 \\
				1 + \lambda
			\end{bsmallmatrix},
			\bm \beta = \begin{bsmallmatrix}
				0 \\
				\lambda \\
				\lambda^{2}
			\end{bsmallmatrix}.
		\]
		问 $\lambda$ 取何值时，有：
		\begin{enumerate}
			\item $\beta$ 可由 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{3}$ 线性表出，且表达式唯一;
			\item $\beta$ 可由 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{3}$ 线性表出，但表达式不唯一;
			\item $\beta$ 不能由 $\bm \alpha_{1},\bm \alpha_{2},\bm \alpha_{3}$ 线性表出.
		\end{enumerate}
	\end{titwo}

	\begin{titwo}
		已知 $\bm \alpha_{1}, \bm \alpha_{2}, \cdots, \bm \alpha_{s}$ 线性无关，$\beta$ 可由 $\bm \alpha_{1}, \bm \alpha_{2},$ $\cdots, \bm \alpha_{s}$ 线性表出，且表达式的系数全不为零. 证明：$\bm \alpha_{1}, \bm \alpha_{2}, \cdots, \bm \alpha_{s},\beta$ 中任意 $s$ 个向量均线性无关.
	\end{titwo}